[LeetCode]Linked List Cycle II

题目描述：

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Follow up:
Can you solve it without using extra space?

解题方案：

假设循环节点距离头节点距离为L，那么当快指针和慢指针第一次相遇时，快指针和慢指针走的距离分别是K1, K2，则：

K1 = 2K2, K1 - K2 = (S - L)的整数倍，那么K2 = S - L。

下面让慢指针回到头节点，快指针不动，则快慢指针距离循环结点的距离都是L，所以当快慢节点每次都增加1的时候，相遇点就是循环点。下面是该题代码：

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        int IsCircle = 0;
        struct ListNode *slow = head;
        struct ListNode *fast = head;

        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;

            if (slow == fast) {
                IsCircle = 1;
                break;
            }
        }

        if (IsCircle == 0) {
            return NULL;
        }

        slow = head;
        while(fast != slow) {
            fast = fast->next;
            slow = slow->next;
        }

        return fast;
    }
};